Step 1: Understanding the Topic:
This question belongs to "Chemical Kinetics." It explores the units of the rate constant ($k$), which is a fundamental part of the rate law for a chemical reaction. The units of $k$ are not fixed; they change depending on the overall order of the reaction. This ensures that the overall "Rate" of the reaction always maintains consistent units of concentration divided by time ($mol \cdot L^{-1} \cdot s^{-1}$).
Step 2: Key Formulas and Approach:
The general formula for determining the units of the rate constant for a reaction of order $n$ is:
\[ \text{Units of } k = (mol \cdot L^{-1})^{1-n} \cdot s^{-1} \]
We can also write this as $M^{1-n} \cdot s^{-1}$ where $M$ is molarity. By substituting the values $n = 0, 1, 2, 3$, we find the corresponding units.
Step 3: Detailed Explanation:
A. Zero order (n=0): Substituting $n=0$: $(mol \cdot L^{-1})^{1-0} \cdot s^{-1} = mol \cdot L^{-1} \cdot s^{-1}$. Matches with IV.
B. First order (n=1): Substituting $n=1$: $(mol \cdot L^{-1})^{1-1} \cdot s^{-1} = (mol \cdot L^{-1})^0 \cdot s^{-1} = s^{-1}$. Matches with III.
C. Second order (n=2): Substituting $n=2$: $(mol \cdot L^{-1})^{1-2} \cdot s^{-1} = (mol \cdot L^{-1})^{-1} \cdot s^{-1} = L \cdot mol^{-1} \cdot s^{-1}$. Matches with I.
D. Third order (n=3): Substituting $n=3$: $(mol \cdot L^{-1})^{1-3} \cdot s^{-1} = (mol \cdot L^{-1})^{-2} \cdot s^{-1} = L^2 \cdot mol^{-2} \cdot s^{-1}$. Matches with II.
Step 4: Final Answer:
The correct matching sequence is A-IV, B-III, C-I, D-II, which is option (B).